In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B(see Fig. 7.23). Show that:
(i) ∆ AMC ≅ ∆ BMD
(ii) ∠ DBC is a right angle.
(iii) ∆ DBC ≅ ∆ ACB
(iv) CM = $\frac{1}{2}$ AB

 

Given: M is the mid-point of hypotenuse AB, ∠C = 90° and DM = CM

To Prove:

i) ΔAMC ≅ ΔBMD

ii) ∠DBC is a right angle .

iii) ΔDBC ≅ ΔACB

iv) CM = $\frac{1}{2}$  AB

Proof: i) In ΔAMC and ΔBMD,

AM = BM (M is the mid - point of AB)

∠AMC = ∠BMD (Vertically opposite angles)

CM = DM (Given)

∴ ΔAMC ≅ ΔBMD (By SAS congruence rule)

∴ AC = BD (By CPCT)

Also, ∠ACM = ∠BDM (By CPCT)

ii) ∠DBC is a right angle.

We know that, ∠ACM = ∠BDM (proved above)

But, ∠ACM and ∠BDM are alternate interior angles. Since alternate angles are equal, it can be said that DB || AC.

∠DBC + ∠ACB = 180° (Co-interior angles)

∠DBC + 90° = 180° [Since, ΔACB is a right angled triangle]

∴ ∠DBC = 90°

Thus, ∠DBC is a right angle.

iii) In ΔDBC and ΔACB,

DB = AC (Already proved)

∠DBC = ∠ACB = 90° (Proved above)

BC = CB(Common)

∴ Δ DBC ≅ Δ ACB (SAS congruence rule)

iv) CM = 1/2 AB

Since Δ DBC ≅ Δ ACB

AB = DC (By CPCT)

⇒ 1/2 AB = 1/2 DC

It is given that M is the midpoint of DC and AB.

CM = 1/2 DC = 1/2 AB

∴ CM = 1/2 AB